Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength `6.2 xx 10^(-6)m` on a metal surface. The work function of the metal is 0.1 eV ?
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If a light with frequency `4xx10^(16) Hz` emitted photoelectrons with double the maximum kinetic as are emitted by the light of frequency `2.5xx10^(16
Correct Answer - `1xx10^(16) Hz` For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)` `:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)` For frequency `4xx10^(16)` Hz, `h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)` Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....
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When a certain metal was irradiated with light frequency `1.6xx10^(16) Hz`, the photo electrons emitted had twice the kinetic energy as did photoelect
Correct Answer - 4 By photoelectric effect `hv=hv_(0)+KE` `:. KE_(1)=h(v_(1)-v_(0)) …(1)` `KE_(2)=h(v_(2)-v_(0))=KE_(1)//2 …(2)` Dividing equation (2) by (1) we have `(v^(2)-v_(0))/(v_(1)-v_(0))=1/2` `(1.0xx10^(16))/(1.6xx10^(16)-v_(0))=1/2 2.0 xx10^(16)-2v_(0)=1.6xx10^(16)-v_(0)` `v_(0)=4xx10^(15) Hz`
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The work function for a metal is `40 eV`. To emit photo electrons of zero velocity from the surface of the metal the wavelength of incident light shou
Correct Answer - 31 `W_(0)+K=hv` `4xx1.6xx10^(-19)=(6.62xx10^(-34)xx3xx10^(8))/(lambda)` `lambda=3.18xx10^(-7)=31xx10^(-8) m` `lambda=31 nm`
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The threshold wavelength `(lambda_(0))` of sodium metal is `6500 Å`. If UV light of wavelength `360 Å` is used, what will be kinetic energy of the pho
Correct Answer - 3 The threshold frequency `(v_(0))` corresponding to the wavelength `6500 Å` is `c//lambda_(0)` Therefore, the threshold energy `=hv_(0)hc//lambda_(0)` Substituting for `h, c` and `lambda_(0)` we gwt threshold energy...
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Light of wavelength `lambda` strikes a metal surface with intensity `X` and the metal emits `Y` electrons per second of average energy `Z`. What will
Correct Answer - 1 Number of emitted electron `prop` Intensity of incident light.
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For silver metal, threshold frequency is `1.1 xx 10^(17) s^(-1)`. What is the maximum kinetic energy of the photoelectrons prodcued by shining ultravi
Correct Answer - A `E_("subjected")=hv_(0)+KE` `thereforeKE = 6.6 xx10^(-34)xx[(3xx10^(8))/(1.5 xx 10^(-9))-1.1xx10^(17)]` `= 5.94 xx10^(-17)J`
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When photons of wavelength `lambda_(1) = 2920 Å` strike the surface of metal A, the ejected photoelectron have maximum kinetic energy of `k_(1) eV` an
Correct Answer - (a) `phi_(A) = 2.25 eV, phi_(B) = 4.2 eV` (b) `2e V`
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Photoelectric work function for a surface is 2.4 eV. Light of wavelength `6.8 xx 10^(-7)`m shines on the surface. Find the frequency of incident light
Correct Answer - `v =4.41 xx 10^(14)Hz,v_(0)=5.82 xx 10^(14) Hz`, no photoelectric emission occurs as `v lt v_(0)`
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The photoelectric threshold wavelength for lithium is `8000 Å`. Find the maximum kinetic energy in eV of the emitted electrons from the surface by lig
Correct Answer - 0.5156 eV
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The energy of photoelectrons emitted from a sensitive plate is 1.56 eV. If its threshold old wavelength is `2500 Å`, calculate the wavelength of incid
Correct Answer - `1379 Å`
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